MHTCET 16 Apr 2026 Shift 2 Paper

Concept:Surface energy of a drop is directly proportional to its surface area.When small drops merge, total volume remains conserved.Explanation:Let the radius of each small drop be $r$.Surface area of one small drop is $4 \pi r^2$.For 1000 small drops, initial total surface area is$$1000 \times 4 \pi r^2 = 4000 \pi r^2.$$Let the radius of the big drop be $R$.Since volume is conserved,$$1000 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3.$$Cancelling $\frac{4}{3} \pi$ from both sides,$$1000 r^3 = R^3.$$Taking cube root,$$R = 10 r.$$The final surface area of the big drop is$$4 \pi R^2 = 4 \pi (10 r)^2 = 400 \pi r^2.$$Since surface energy is proportional to surface area,$$\frac{\text{Final surface energy}}{\text{Initial surface energy}} = \frac{400 \pi r^2}{4000 \pi r^2} = \frac{1}{10}.$$Therefore, the required ratio is $1 : 10$.Answer:The ratio is $1 : 10$.Correct option: B.